Find q when 22.0 g of water is heated from 45.0°C to 100.0°C.?

1 Answer

  • 4 weeks ago
    Favorite Answer

    We know the relationship

    q = m*s*∆T

    Where q is heat, m is mass of substance, s is specific heat

    ∆T is change in temperature

    From Question,

    m = 22.0 g

    specific heat of water, s = 4.184 Jg-1oC-1

    ∆T = (100-45)oC = 55oC

    q = 22.0 g *4.184 Jg-1oC-1 *55oC

    q = 5062.64 J

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